2002 AMC 10B Problem 3

The mean is the total divided by $9$. Every number in the set is made of nines, and dividing by $9$ is easy for those, so divide first and add afterwards.

1. Notice $9 = 9 \times 1$, $99 = 9 \times 11$, $999 = 9 \times 111$, and so on. A number made of $k$ nines is $9$ times the number made of $k$ ones.
2. So the total is $9 \times (1 + 11 + 111 + \cdots + 111111111)$, and dividing by $9$ gives
$$M = 1 + 11 + 111 + \cdots + 111111111.$$
3. Add this column by column. Line the numbers up on the right:

Every one of the nine numbers has a ones digit, eight of them have a tens digit, and so on, until only the longest number reaches the ninth column.
4. No column adds to more than $9$, so nothing carries. Reading the columns from left to right gives
$$M = 123456789.$$
5. The digits are $1$ through $9$. The only digit missing is $0$.

The answer is $\boxed{\textbf{(A)}\ 0}$.

Numbers like $999\ldots9$ are almost always easier to handle as $9 \times 111\ldots1$ (or as a power of ten minus one). Here, dividing by $9$ before adding turned a messy sum into stacked rows of ones, and those add with no carrying. Next time you see all-nines numbers, try one of these two rewrites first.

Each number is a power of ten minus one: $9 = 10 - 1$, $99 = 100 - 1$, and so on up to $10^9 - 1$. The powers of ten add to $1111111110$, and there are nine "minus ones", so the total is $1111111110 - 9 = 1111111101$. Long division by $9$ gives $123456789$, which has no $0$.